Linear Algebra Examples

$5 x - 4 y = - 3$ , $8 x - y = 2$

Write the system of equations in matrix form.

$[\begin{array}{c} 5 & - 4 & - 3 \\ 8 & - 1 & 2 \end{array}]$

Row reduce.

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Perform the row operation $R_{1} = \frac{1}{5} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{5} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} \\ 8 & - 1 & 2 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{5} R_{1}$ .

$[\begin{array}{c} (\frac{1}{5}) \cdot (5) & (\frac{1}{5}) \cdot (- 4) & (\frac{1}{5}) \cdot (- 3) \\ 8 & - 1 & 2 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 8 & - 1 & 2 \end{array}]$

Perform the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 8 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ (- 8) \cdot (1) + 8 & (- 8) \cdot (- \frac{4}{5}) - 1 & (- 8) \cdot (- \frac{3}{5}) + 2 \end{array}]$

$R_{2} = - 8 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & \frac{27}{5} & \frac{34}{5} \end{array}]$

Perform the row operation $R_{2} = \frac{5}{27} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{5}{27} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ \frac{5}{27} R_{2} & \frac{5}{27} R_{2} & \frac{5}{27} R_{2} \end{array}]$

$R_{2} = \frac{5}{27} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{5}{27} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ (\frac{5}{27}) \cdot (0) & (\frac{5}{27}) \cdot (\frac{27}{5}) & (\frac{5}{27}) \cdot (\frac{34}{5}) \end{array}]$

$R_{2} = \frac{5}{27} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Perform the row operation $R_{1} = \frac{4}{5} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{4}{5} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{4}{5} R_{2} + R_{1} & \frac{4}{5} R_{2} + R_{1} & \frac{4}{5} R_{2} + R_{1} \\ 0 & 1 & \frac{34}{27} \end{array}]$

$R_{1} = \frac{4}{5} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{4}{5} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{4}{5}) \cdot (0) + 1 & (\frac{4}{5}) \cdot (1) - \frac{4}{5} & (\frac{4}{5}) \cdot (\frac{34}{27}) - \frac{3}{5} \\ 0 & 1 & \frac{34}{27} \end{array}]$

$R_{1} = \frac{4}{5} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & \frac{11}{27} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x = \frac{11}{27}$

$y = \frac{34}{27}$

The solution is the set of ordered pairs that makes the system true.

$(\frac{11}{27}, \frac{34}{27})$

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